Question

At point A on the diameter AB of a circle of radius 10 cm, tangent XAY is drawn to the circle. Find the length of the chord CD parallel to XY at a distance of 16 cm from A.

Answer 1

\[ \begin{array}{r l} \textbf{Given:} & \text{A circle with centre } O \text{ and radius } 10 \text{ cm, tangent } XAY \text{ at the end } A \text{ of the diameter } AB, \text{ and a chord } CD \text{ parallel to } XY \text{ at a distance } 16 \text{ cm from } A. \\[4pt] \textbf{To Find:} & \text{The length of the chord } CD. \\[4pt] \textbf{Solution:} & \text{Let } CD \text{ meet } AB \text{ at } M, \text{ so that } AM = 16 \text{ cm and } OA = OC = 10 \text{ cm.} \\[4pt] & \text{The radius through the point of contact is perpendicular to the tangent, so } AB \text{ is perpendicular to } XY, \text{ and since } CD \text{ is parallel to } XY, \text{ } OM \text{ is perpendicular to } CD. \\[4pt] & \text{The perpendicular from the centre to a chord bisects the chord, hence } CD = 2\,CM. \\[4pt] & \text{In the right triangle } OMC, \text{ by Pythagoras' theorem:} \\[4pt] & \begin{aligned} OM &= AM - OA \\[4pt] &= 16 - 10 \\[4pt] &= 6 \\[4pt] CM^{2} &= OC^{2} - OM^{2} \\[4pt] &= 10^{2} - 6^{2} \\[4pt] &= 100 - 36 \\[4pt] &= 64 \\[4pt] CM &= 8 \\[4pt] CD &= 2 \times 8 \\[4pt] &= 16 \end{aligned} \\[4pt] \textbf{Answer:} & CD = 16 \text{ cm} = 16.00 \text{ cm.} \end{array} \]